Chapter 4: Clustering and classification
Marc Denojean-Mairet
27/11/2021
Chapter 4: Clustering and Classification
Part 2: explore the structure and the dimensions of the data
Loading the Boston data.
# set plots size
knitr::opts_chunk$set(fig.width=16, fig.height=10)
#code from DataCamp.
#access the MASS package
library (dplyr)
library(MASS)
library(corrplot)
library(tidyr)
#load the data
data("Boston")
#explore the dataset
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...dim(Boston)## [1] 506 14summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00Description:
The Boston data set represent the housing values in Suburbs of Boston.
The Boston data frame has 506 rows and 14 columns.
This data frame contains the following columns:
crim = per capita crime rate by town
zn = proportion of residential land zoned for lots over 25,000 sq.ft
indus = proportion of non-retail business acres per town
chas = Charles River dummy variable (= 1 if tract bounds river; 0 otherwise)
nox = nitrogen oxides concentration (parts per 10 million)
rm = average number of rooms per dwelling
age = proportion of owner-occupied units built prior to 1940
dis = weighted mean of distances to five Boston employment centres
rad = index of accessibility to radial highways
tax = full-value property-tax rate per $10,000
ptratio = pupil-teacher ratio by town
black = 1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town
lstat = lower status of the population (percent)
medv = median value of owner-occupied homes in $1000s
Source:
Harrison, D. and Rubinfeld, D.L. (1978) Hedonic prices and the demand for clean air. J. Environ. Economics and Management 5, 81–102.
Belsley D.A., Kuh, E. and Welsch, R.E. (1980) Regression Diagnostics. Identifying Influential Data and Sources of Collinearity. New York: Wiley.
Part 3: graphical overview of the data
#plot matrix of the variables
pairs(Boston, gap=1/30)
# MASS, corrplot, tidyr and Boston dataset are available
# calculate the correlation matrix and round it
cor_matrix<-cor(Boston) %>% round(digits = 2)
# print the correlation matrix
cor_matrix## crim zn indus chas nox rm age dis rad tax ptratio
## crim 1.00 -0.20 0.41 -0.06 0.42 -0.22 0.35 -0.38 0.63 0.58 0.29
## zn -0.20 1.00 -0.53 -0.04 -0.52 0.31 -0.57 0.66 -0.31 -0.31 -0.39
## indus 0.41 -0.53 1.00 0.06 0.76 -0.39 0.64 -0.71 0.60 0.72 0.38
## chas -0.06 -0.04 0.06 1.00 0.09 0.09 0.09 -0.10 -0.01 -0.04 -0.12
## nox 0.42 -0.52 0.76 0.09 1.00 -0.30 0.73 -0.77 0.61 0.67 0.19
## rm -0.22 0.31 -0.39 0.09 -0.30 1.00 -0.24 0.21 -0.21 -0.29 -0.36
## age 0.35 -0.57 0.64 0.09 0.73 -0.24 1.00 -0.75 0.46 0.51 0.26
## dis -0.38 0.66 -0.71 -0.10 -0.77 0.21 -0.75 1.00 -0.49 -0.53 -0.23
## rad 0.63 -0.31 0.60 -0.01 0.61 -0.21 0.46 -0.49 1.00 0.91 0.46
## tax 0.58 -0.31 0.72 -0.04 0.67 -0.29 0.51 -0.53 0.91 1.00 0.46
## ptratio 0.29 -0.39 0.38 -0.12 0.19 -0.36 0.26 -0.23 0.46 0.46 1.00
## black -0.39 0.18 -0.36 0.05 -0.38 0.13 -0.27 0.29 -0.44 -0.44 -0.18
## lstat 0.46 -0.41 0.60 -0.05 0.59 -0.61 0.60 -0.50 0.49 0.54 0.37
## medv -0.39 0.36 -0.48 0.18 -0.43 0.70 -0.38 0.25 -0.38 -0.47 -0.51
## black lstat medv
## crim -0.39 0.46 -0.39
## zn 0.18 -0.41 0.36
## indus -0.36 0.60 -0.48
## chas 0.05 -0.05 0.18
## nox -0.38 0.59 -0.43
## rm 0.13 -0.61 0.70
## age -0.27 0.60 -0.38
## dis 0.29 -0.50 0.25
## rad -0.44 0.49 -0.38
## tax -0.44 0.54 -0.47
## ptratio -0.18 0.37 -0.51
## black 1.00 -0.37 0.33
## lstat -0.37 1.00 -0.74
## medv 0.33 -0.74 1.00# visualize the correlation matrix
corrplot(cor_matrix, method="circle", type="upper", cl.pos="b", tl.pos="d", tl.cex = 0.6)
corrplot(cor_matrix, method = 'number', type="upper")
Looking at the data, we see that medv has a positive correlation with rm and a negative correlation with lstat. This make sense as a house would have a higher price based on the number of rooms. As well, house located in a lower status population area would have a lower price. We also see that nox has a positive correlation with indus and age. nox has a negative correlation with dis. The higher the number of industry, the higher the emission of nitrogen oxides. Older houses would probably be mostly in old industrial areas of the city. medv has a positive correlation with crim but only at 0.33. At first glance, the value of houses is mainly due to th e number of rooms per dwelling. Taxes has a small negative correlation with medv.
Part 4: standardize the dataset
During scaling of the data, the mean is subtracted from the column and the difference is divided by the standard deviation. This is an example of the data before and after scaling.
Before:
row 1: rm = 6.575
After:
row 1: rm = 0.413262920
# center and standardize variables
boston_scaled <- scale(Boston)
# summaries of the scaled variables
summary(boston_scaled)## crim zn indus chas
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563 Min. :-0.2723
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723
## Median :-0.390280 Median :-0.48724 Median :-0.2109 Median :-0.2723
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202 Max. : 3.6648
## nox rm age dis
## Min. :-1.4644 Min. :-3.8764 Min. :-2.3331 Min. :-1.2658
## 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049
## Median :-0.1441 Median :-0.1084 Median : 0.3171 Median :-0.2790
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617
## Max. : 2.7296 Max. : 3.5515 Max. : 1.1164 Max. : 3.9566
## rad tax ptratio black
## Min. :-0.9819 Min. :-1.3127 Min. :-2.7047 Min. :-3.9033
## 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049
## Median :-0.5225 Median :-0.4642 Median : 0.2746 Median : 0.3808
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332
## Max. : 1.6596 Max. : 1.7964 Max. : 1.6372 Max. : 0.4406
## lstat medv
## Min. :-1.5296 Min. :-1.9063
## 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 3.5453 Max. : 2.9865# class of the boston_scaled object
class(boston_scaled)## [1] "matrix" "array"# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)Use the quantiles as the break points in the categorical variable and divide the dataset to train and test sets
# summary of the scaled crime rate
summary(boston_scaled$crim)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -0.419367 -0.410563 -0.390280 0.000000 0.007389 9.924110# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
bins## 0% 25% 50% 75% 100%
## -0.419366929 -0.410563278 -0.390280295 0.007389247 9.924109610# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))
# look at the table of the new factor crime
table(crime)## crime
## low med_low med_high high
## 127 126 126 127# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
# number of rows in the Boston dataset
n <- nrow(boston_scaled)
# choose randomly 80% of the rows
ind <- sample(n, size = n * 0.8)
# create train set
train <- boston_scaled[ind,]
# create test set
test <- boston_scaled[-ind,]
# save the correct classes from test data
correct_classes <- test$crime
# remove the crime variable from test data
test <- dplyr::select(test, -crime)Now, 80% of the data belongs to the train set. Saved the crime categories from the test set and removed the categorical crime variable from the test dataset.
Part 5: linear discriminant analysis
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
# print the lda.fit object
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2326733 0.2599010 0.2574257 0.2500000
##
## Group means:
## zn indus chas nox rm age
## low 0.9420737 -0.9100351 -0.1466769325 -0.8779834 0.39012753 -0.8862989
## med_low -0.1154342 -0.2588118 0.0276404741 -0.5440944 -0.11821937 -0.3753394
## med_high -0.3866438 0.1822975 0.1062382627 0.3452508 0.09497491 0.3621578
## high -0.4872402 1.0171306 0.0005392655 1.0518773 -0.46891266 0.8092531
## dis rad tax ptratio black lstat
## low 0.9370073 -0.6849803 -0.7416949 -0.43497806 0.37824493 -0.74897671
## med_low 0.2846338 -0.5585791 -0.4739892 -0.07184181 0.32146978 -0.18736929
## med_high -0.3503424 -0.3888647 -0.2922584 -0.23128799 0.05194819 0.04768121
## high -0.8490243 1.6379981 1.5139626 0.78062517 -0.73848267 0.83917220
## medv
## low 0.43852403
## med_low 0.02625602
## med_high 0.18033283
## high -0.63290848
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.126575584 0.792867221 -0.8481244
## indus 0.005201058 -0.228673416 0.2843291
## chas -0.075078528 -0.008113732 0.1798857
## nox 0.381671219 -0.652168067 -1.4957761
## rm -0.137186163 -0.121609726 -0.2138480
## age 0.304423457 -0.188827681 -0.2310376
## dis -0.040766424 -0.207189069 -0.1558653
## rad 3.055104822 1.034843173 -0.1331417
## tax -0.054543581 -0.145360642 0.7124490
## ptratio 0.135527867 -0.010757572 -0.2762576
## black -0.147054901 0.046438934 0.1365084
## lstat 0.242166019 -0.390852459 0.1614414
## medv 0.231479519 -0.520393302 -0.3194708
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9470 0.0399 0.0131# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)
Part 6: Save the crime categories
# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)## predicted
## correct low med_low med_high high
## low 19 11 3 0
## med_low 5 11 5 0
## med_high 0 6 16 0
## high 0 0 0 26The best predictions are for high. Also, the model does not predict very well med_high. We can see in the LD plot that med_high and med_low are in the same side. The model predicts high correctly because the distance between high and low, med_low, and med_high is large.
Part 7: Reload the Boston dataset and standardize the dataset
# load MASS and Boston
library(MASS)
data('Boston')
# scale data
Boston = as.data.frame(scale(Boston))
# euclidean distance matrix
dist_eu <- dist(Boston)
# look at the summary of the distances
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970# manhattan distance matrix
dist_man <- dist(Boston, method = 'manhattan')
# look at the summary of the distances
summary(dist_man)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.2662 8.4832 12.6090 13.5488 17.7568 48.8618cluster with centers = 3
# k-means clustering
km <-kmeans(Boston, centers = 3)
# plot the Boston dataset with clusters
pairs(Boston[6:10], col = km$cluster)
cluster with centers = 1
# k-means clustering
km <-kmeans(Boston, centers = 1)
# plot the Boston dataset with clusters
pairs(Boston[6:10], col = km$cluster)
cluster with centers = 2
# k-means clustering
km <-kmeans(Boston, centers = 2)
# plot the Boston dataset with clusters
pairs(Boston[6:10], col = km$cluster)
The cluster with centers = 3 seems to be the best.
Bonus and super bonus
library(ggplot2)
# Boston dataset is available
set.seed(123)
# determine the number of clusters
k_max <- 10
# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(Boston, k)$tot.withinss})
# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')
# k-means clustering
km <-kmeans(Boston, centers = 2)
# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
#install.packages("plotly")
library(plotly)
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')